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dexar [7]
3 years ago
10

A customer works into a post office to purchase stamps. she purchases a total of 45 stamps. some of the stamps are 49cent stamps

and some are 33cents stamps. how many 49cent stamps and how many 33cent stamps did she purchase if the total cost of all the stamps was $17.89
Mathematics
1 answer:
baherus [9]3 years ago
5 0

Step-by-step explanation:Let number of 49 cents =A

                                             Let number of 33 cents=B

<u>A+B=45    equation 1</u>

1 $=100 cents

49 cents =.49 dollar

33 cents=.33 dollar

so .<u>49A+.33B=17.89    equation 2</u>

Add equation 1 and equation 2

since these are simultaneous equations

.49A+.33B=17.89

     A+     B=45

so divide equation 2 by 100 we get

49A+33B=1789

    A+B     =45 multiply equation 1 by 33 on both the sides we get

<u>33A+33B=1485     equation 3</u>

so subtract equation 3 from equation 2 we get

49A+33B=1789  equation 2

<u>33A+33B=1485    equation 3</u>

<u>16A+0      =304</u>

16A=304

A=304/16

A=19

putting nalue of A in equation 1 we get

A+B=45

19+B=45

B=45-19

B=26

so number of 49 cent stamps are 19

and number of 33 cent stamps are 26

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algol13

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3 years ago
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Lorico [155]

Answer:

-\frac{3\sqrt[3]{t} }{2}

Step-by-step explanation:

1: Write g(t) as y, resulting in y=-\frac{8}{27}t^3

2: Interchange the variables y and t, resulting in t=-\frac{8}{27}y^3

3: Multiply both sides by 27, resulting in 27t=-8y^3

4: Divide both sides by -8, resulting in -\frac{27t}{8}=y^3

5: Find the cube root of both sides, resulting in \sqrt[3]{-\frac{27t}{8} }=y

6: Apply a radical rule, resulting in -\sqrt[3]{\frac{27t}{8} } =y

7: Apply another radical rule, resulting in -\frac{\sqrt[3]{27t} }{\sqrt[3]{8} } =y

8: Simplify the denominator, resulting in  -\frac{\sqrt[3]{27t} }{2} =y

9: Apply yet another radical rule, resulting in -\frac{\sqrt[3]{27}\sqrt[3]{t}   }{2} =y

10: Simplify \sqrt[3]{27}, resulting in -\frac{3\sqrt[3]{t}   }{2} =y

3 0
3 years ago
The combined age of three relatives is 120 years. James is three times the age of Dan, and Paul is two time the sum of the ages
BaLLatris [955]

Answer:

Step-by-step explanation:

Let x represent Dan's age

3x be James'age

Paul 2(3x+x)=6x+2x

Suming all ages

x+3x+6x+2x=120

Solving for x

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8 0
4 years ago
1. Find the product. Simplify. 13/14 of 10/9. A. 15/42 B. 5/21 C. 2/7 D. 5/6 2. Andy has 5 pots, and each pot can hold 3/8 pound
Rufina [12.5K]

1) Resultant fraction: \frac{65}{63}

2) Total soil: 1\frac{7}{8} pounds

3) Result of the product: \frac{35}{2}

Step-by-step explanation:

1)

In this problem, we want to find the product between the two fractions

\frac{13}{14}

and

\frac{10}{9}

In order to find this product, we have to multiply the numerators of each fraction and the denominators of each fraction. We get:

\frac{13}{14}\cdot \frac{10}{9}=\frac{13\cdot 10}{14\cdot 9}=\frac{130}{126}

Now we simplify, dividing both numerator and denominator by 2:

\frac{130/2}{126/2}=\frac{65}{63}

And the fraction cannot be further simplified.

2)

Here we have:

n = 5 (number of pots that Andy has)

s=\frac{3}{8} (amount of soil (in pounds) that each pot can contain)

In order to find the amount of soil that Andy needs to fill all the pots, we have to multiply the number of pots (n) by the amount of soil that each pot can contain (s).

If we do so, we find:

t=n \cdot s = 5 \cdot \frac{3}{8}=\frac{5\cdot 3}{8}=\frac{15}{8}

Which can be rewritten as a mixed fraction as:

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3)

Here we want to find the result of the following product:

4\frac{2}{3}\cdot 3\frac{3}{4}

In order to do so, we first have to rewrite each fraction as an improper fraction.

For the 1st fraction:

4\frac{2}{3}=\frac{4\cdot 3+2}{3}=\frac{12+2}{3}=\frac{14}{3}

For the 2nd fraction:

3\frac{3}{4}=\frac{3\cdot 4+3}{4}=\frac{12+3}{4}=\frac{15}{4}

Now we can finally find the product of the two fractions:

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Where in the last step, we divide both the numerator and the denominator by 6.

Learn more about fractions:

brainly.com/question/605571

brainly.com/question/1312102

#LearnwithBrainly

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mario62 [17]

Answer:

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EH = 10

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3 years ago
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